Why Use a Two-Way Sensitivity Table?
A one-at-a-time shock answers a useful but limited question: what changes when one assumption moves and the others stay fixed? Long-term product cash flows can depend on assumptions jointly. For example, mortality changes the probability of a benefit payment, while the interest rate changes the present value of that payment. A two-way table shows the result of changing both assumptions together and allows the paired effect to be compared with the separate effects.
A sensitivity table is only interpretable when the model basis is controlled. Each cell should use the same contract, cash-flow timing, reserve definition, and assumptions other than the two being varied. If a changed assumption also affects a benefit, premium, or expense, that linkage must be stated and recalculated consistently.
The table is not merely a collection of separate one-factor shocks. The response to changing one assumption may depend on the level of the other assumption. A useful diagnostic is to compare the change across one dimension at different values of the other dimension. If those changes differ, the model has an interaction over the tested range. The difference need not imply a complicated model: even a product of a probability and a discounted cash flow can produce it.
Two-Year Reserve Model
Consider a hypothetical two-year term contract. A benefit \(B\) is payable at the end of each year in which death occurs. A premium \(P\) is payable at time 1 if the policyholder survives the first year. There are no other cash flows. Annual conditional mortality is \(q\) in both years, and the effective annual valuation interest rate is \(i\), with \(v=1/(1+i)\). Mortality is independent from year to year conditional on being in force, and the same \(q\) applies in each year.
Define \(V_0\) as the prospective reserve immediately after the time-0 premium, conditional on the contract being in force then. The probability of death in year 1 is \(q\); the probability of surviving year 1 and dying in year 2 is \((1-q)q\); and the probability of surviving to time 1, when the future premium is paid, is \(1-q\). Therefore:
This formula is conditional on the policy being in force at time 0. It discounts each end-of-year benefit to time 0 and subtracts the discounted expected value of the future premium. It does not include the time-0 premium, which has already been paid. It also would not apply unchanged if benefits were paid at a different time, premiums were payable on a different schedule, mortality varied by year, or lapse and expenses were included.
Worked Example: Mortality and Interest Sensitivity of a Reserve
For the two-year contract above, let \(B=\$500\) and \(P=\$50\). Compare mortality \(q=0.08\) and \(q=0.12\) with valuation rates \(i=0.03\) and \(i=0.07\). Find the reserve in each cell and interpret the paired changes.
Method. Recalculate the prospective reserve formula for each pair \((q,i)\), changing no other contract feature. The discount factors are \(v=1/1.03=0.970874\) and \(v=1/1.07=0.934579\), to six decimal places. The corresponding \(v^2\) values are \(0.942596\) and \(0.873439\).
For \(q=0.08,\ i=0.03\), the benefit present value is \(500[0.08(0.970874)+(0.92)(0.08)(0.942596)]=\$73.52\). The future premium present value is \(50(0.92)(0.970874)=\$44.66\). Thus \(V_0=73.52-44.66=\$28.86\).
For \(q=0.12,\ i=0.03\), the benefit present value is \(500[0.12(0.970874)+(0.88)(0.12)(0.942596)]=\$108.02\). The future premium present value is \(50(0.88)(0.970874)=\$42.72\). Thus \(V_0=108.02-42.72=\$65.30\).
For \(q=0.08,\ i=0.07\), the benefit present value is \(500[0.08(0.934579)+(0.92)(0.08)(0.873439)]=\$69.53\). The future premium present value is \(50(0.92)(0.934579)=\$42.99\). Thus \(V_0=69.53-42.99=\$26.54\).
For \(q=0.12,\ i=0.07\), the benefit present value is \(500[0.12(0.934579)+(0.88)(0.12)(0.873439)]=\$102.19\). The future premium present value is \(50(0.88)(0.934579)=\$41.12\). Thus \(V_0=102.19-41.12=\$61.07\).
| Annual mortality \(q\) | Valuation rate \(i=0.03\) | Valuation rate \(i=0.07\) |
|---|---|---|
| 0.08 | \$28.86 | \$26.54 |
| 0.12 | \$65.30 | \$61.07 |
The higher mortality produces a higher reserve at either interest rate: it raises the expected value of death benefits and reduces expected future premiums. The higher valuation rate lowers the reserve in both rows because it discounts future cash flows more heavily. The reserve change from \(q=0.08\) to \(q=0.12\) is \(\$36.44\) at \(i=0.03\), versus \(\$34.54\) at \(i=0.07\). The mortality effect therefore differs across the two interest-rate columns; this is an interaction in the table.
Check. Each reserve equals the benefit present value minus the future premium present value, and the higher-mortality cells are larger in this model. A common mistake is to include the time-0 premium in \(V_0\), even though the reserve is defined immediately after it has been paid.
Why Paired Changes Can Interact
The mechanism can be seen directly in the reserve formula. The second-year benefit term contains \((1-q)qv^2\). Mortality appears both as the probability of surviving the first year and as the conditional probability of dying in the second year; interest appears through \(v^2\). Consequently, changing mortality changes the amount of cash flow to which the discount factor applies, while changing interest changes the value of that mortality-weighted cash flow.
State whether each cell is a reserve, expected profit, or another output, and fix the contract terms, timing, and all unshocked assumptions.
Use clearly labelled row and column values. Include the base assumptions if the purpose is to compare shocks with a baseline.
Apply both assumptions together in the model. Do not add two separate one-at-a-time changes unless the model has been shown to be additive.
Compare the output across rows and columns, then ask whether a change in one assumption is different at different levels of the other.
For a four-cell grid with row values \(a_1,a_2\) and column values \(b_1,b_2\), a cross-comparison can be written as \(Y(a_2,b_2)-Y(a_2,b_1)-Y(a_1,b_2)+Y(a_1,b_1)\), where \(Y\) is the output. This compares the change across the second dimension at one level of the first with the corresponding change at the other level. A nonzero result indicates interaction over those selected levels. It is a finite-grid diagnostic, not a claim that the model has the same interaction at every possible assumption value.
In the reserve example, using unrounded cell values, the cross-comparison is approximately \(61.0708-65.3030-26.5351+28.8623=-1.9050\), or \(-\$1.90\) to the nearest cent. The negative sign means the reserve increase associated with the mortality change is smaller at the higher interest rate. It does not mean the reserve itself is negative or that the mortality shock reduces the reserve.
Profit Sensitivity to Mortality and Expenses
A parallel table can be built for expected present-value profit. In the following hypothetical one-year contract, a premium of \(\$100\) and acquisition expense of \(\$10\) occur at time 0. A death benefit of \(\$500\) is paid at time 1 if death occurs during the year. A maintenance expense \(e\) is paid at time 1 only if the policyholder survives. Let annual mortality be \(q\), and use \(i=0.05\). Assume no other cash flows.
The time-0 net inflow is \(100-10=\$90\). The expected time-1 outflow is \(500q+e(1-q)\), so expected present-value profit is:
Worked Example: Paired Mortality and Maintenance-Expense Shocks
Calculate expected present-value profit for \(q=0.04\) and \(q=0.08\), paired with maintenance expenses \(e=\$5\) and \(e=\$10\). Use the contract and 5% effective annual interest rate specified above.
Method. In each cell, weight the death benefit by the probability of death and the maintenance expense by the probability of survival; then discount the expected time-1 outflow. For \(q=0.04,e=\$5\), the expected outflow is \(0.04(500)+0.96(5)=20+4.80=\$24.80\). Its present value is \(24.80/1.05=\$23.6190\), so profit is \(90-23.6190=\$66.38\).
For \(q=0.04,e=\$10\), the expected outflow is \(0.04(500)+0.96(10)=20+9.60=\$29.60\). Its present value is \(29.60/1.05=\$28.1905\), so profit is \(90-28.1905=\$61.81\).
For \(q=0.08,e=\$5\), the expected outflow is \(0.08(500)+0.92(5)=40+4.60=\$44.60\). Its present value is \(44.60/1.05=\$42.4762\), so profit is \(90-42.4762=\$47.52\).
For \(q=0.08,e=\$10\), the expected outflow is \(0.08(500)+0.92(10)=40+9.20=\$49.20\). Its present value is \(49.20/1.05=\$46.8571\), so profit is \(90-46.8571=\$43.14\).
| Annual mortality \(q\) | Expense \(e=\$5\) | Expense \(e=\$10\) |
|---|---|---|
| 0.04 | \$66.38 | \$61.81 |
| 0.08 | \$47.52 | \$43.14 |
Using unrounded profits, the cross-comparison is approximately \(43.1429-47.5238-61.8095+66.3810=\$0.1905\), or \(\$0.19\) to the nearest cent. The expense increase reduces profit by about \(\$4.57\) at \(q=0.04\) and \(\$4.38\) at \(q=0.08\). The expense applies only on survival, so its expected cost depends on mortality. That dependence creates the interaction.
Check. The expected outflow at \(q=0.04,e=\$5\) is \(\$24.80\), less than the \(\$25.00\) obtained by incorrectly treating the expense as certain. That incorrect calculation gives profit \(90-25/1.05=\$66.19\), rather than the correct \(\$66.38\). The correction is to multiply the expense by the survival probability \(1-q\).
Changing Mortality by Duration
A two-way table need not always pair mortality with interest. It can also compare mortality assumptions in different durations. Suppose the first-year conditional mortality is \(q_1\) and second-year conditional mortality is \(q_2\), with the same two-year contract structure as before. The probability of dying in year 2 is now \((1-q_1)q_2\). Thus, at a fixed interest rate:
This is not the equal-mortality formula unless \(q_1=q_2\). Varying \(q_1\) changes both the probability of a first-year claim and the number of policies remaining to pay a second-year claim or the time-1 premium. Varying \(q_2\) changes the second-year claim probability among those who survived year 1.
Worked Example: First- and Second-Year Mortality Sensitivity
Use \(B=\$500\), \(P=\$50\), and \(i=0.05\). Compare \(q_1=0.08,0.12\) with \(q_2=0.06,0.10\). Find the four reserves, rounding only final monetary values to the nearest cent.
Here \(v=1/1.05=0.952381\) and \(v^2=0.907029\). For \(q_1=0.08,q_2=0.06\), the probability-weighted benefit factors are \(0.08v=0.0761905\) and \(0.92(0.06)v^2=0.0500680\). Benefit present value is \(500(0.1262585)=\$63.13\); premium present value is \(50(0.92)(0.952381)=\$43.81\). The reserve is \(\$19.32\).
For \(q_1=0.08,q_2=0.10\), the benefit factors are \(0.0761905\) and \(0.92(0.10)(0.907029)=0.0834467\). Benefit present value is \(500(0.1596372)=\$79.82\), while premium present value remains \(\$43.81\). The reserve is \(\$36.01\).
For \(q_1=0.12,q_2=0.06\), the benefit factors are \(0.12v=0.1142857\) and \(0.88(0.06)v^2=0.0478912\). Benefit present value is \(500(0.1621769)=\$81.09\); premium present value is \(50(0.88)(0.952381)=\$41.90\). The reserve is \(\$39.18\).
For \(q_1=0.12,q_2=0.10\), the benefit factors are \(0.1142857\) and \(0.88(0.10)(0.907029)=0.0798186\). Benefit present value is \(500(0.1941043)=\$97.05\); premium present value is \(\$41.90\). The reserve is \(\$55.15\).
| First-year mortality \(q_1\) | Second-year mortality \(q_2=0.06\) | Second-year mortality \(q_2=0.10\) |
|---|---|---|
| 0.08 | \$19.32 | \$36.01 |
| 0.12 | \$39.18 | \$55.15 |
The second-year mortality increase raises the reserve by about \(\$16.69\) when \(q_1=0.08\), and by about \(\$15.96\) when \(q_1=0.12\). The difference arises because fewer contracts survive the first year at the higher \(q_1\), so fewer can generate a second-year claim. This check confirms that \(q_2\) must be applied conditionally to first-year survivors, not directly to the original in-force group.
Reading Tables Without Overstating Them
A table summarizes the model under specified assumptions; it does not by itself establish which assumptions are most plausible or what action should be taken. Interpret the direction and size of the cell changes in the context of the cash flows. A reserve may increase with mortality because expected claims rise and future premiums fall, while an expected profit may decrease. The sign depends on the output definition and the contract cash flows.
- Keep the basis consistent. Do not change benefit timing, expense timing, or reserve definition between cells.
- Recalculate dependent quantities. If an expense is survival-contingent or a premium is payable only while active, apply the scenario probabilities to it.
- Separate main effects from paired behavior. Do not assume two one-at-a-time changes add to the result of changing both assumptions.
- Label the table precisely. Identify the assumption represented by each axis and state whether rates are annual, conditional, and effective.
These practices are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus. The key exam discipline is to define the output and timing before calculating, then interpret each cell through the model’s actual probability-weighted cash flows.
Check Your Understanding
For each problem, state the relevant cash-flow weighting before giving the requested result.
- A hypothetical one-year policy has a time-0 premium of \(\$80\), acquisition expense of \(\$8\), a time-1 death benefit of \(\$300\), and a time-1 maintenance expense of \(\$4\) only on survival. If \(q=0.05\) and \(i=0.04\), find expected present-value profit. (Answer: \(\$53.92\))
- A two-year contract has \(B=\$200\), \(P=\$20\), equal annual conditional mortality \(q=0.10\), and \(i=0\). Find the reserve immediately after the time-0 premium. (Answer: \(\$20.00\))
- For the one-year profit model in this tutorial, compare \(q=0.04,0.08\) and \(e=\$5,\$10\), with all other assumptions unchanged. What is the cross-comparison of the four profit cells, rounded to the nearest cent? (Answer: \(\$0.19\))
- A two-year contract has \(B=\$600\), \(P=\$30\), \(q_1=0.10\), \(q_2=0.05\), and \(i=0\). Find its time-0 reserve, with the premium at time 1 payable only on survival to that time. (Answer: \(\$60.00\))
- In a reserve table, the increase from a low to a high mortality assumption is \(\$24\) at the low interest rate and \(\$19\) at the high interest rate. Is there an interaction over these tested levels, and what is the difference between the mortality effects? (Answer: Yes; the mortality effect is \(\$5\) smaller at the high interest rate.)